The starship enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. To the crew’s shocking surprise, a Klingon ship is 100 km directly ahead, traveling in the same direction at a measly 20 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over 3.0 s. The Enterprise’s computers react instantly to brake the ship. What acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant. By the way, if it seems familiar it is on this thread, he must have had same textbook, https://www.physicsforums.com/archive/index.php/t-91330.html roffle so d(klingon)= vt + 100 d(enterprise)= .5(vf - vi)t + vi(t) (I got from area under graph) they are equal I thought when they stopped so I combined them: vk(t) + 100 = .5(vf-vi)t + vi(t) vk(t) + 100 = .5vi(t) 20t + 100 = 25t t = 20 s since I hope I have proven my attempt I ended up getting an accerlation of -2500 m/s^2 I am sort of confident I didn't make a mathematical error since I checked it a lot, so I assume I made a conceptual error. Answer in back is -4500 m/s^2. I must have underestimated the question? Thanks for any help you might have to direct me and I am sorry I do not know how to make the equations fancy.